# Difficulty solving &#x222C;<!-- ∬ --> ln &#x2061;<!-- ⁡ --> ( x + y 2 </ms

Difficulty solving $\iint \mathrm{ln}\left(x+{y}^{2}\right)\cdot {y}^{2}\phantom{\rule{thinmathspace}{0ex}}dx\phantom{\rule{thinmathspace}{0ex}}dy$ with change of variables
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Pranav Greer
Let me try to guess. You are interested in a function f:
$\frac{{\mathrm{\partial }}^{2}}{\mathrm{\partial }x\mathrm{\partial }y}f\left(x,y\right)={y}^{2}\mathrm{ln}\left(x+{y}^{2}\right).$
You attempt to do a substitution $u=x+{y}^{2}$ and find function $g\left(u\left(x,y\right),v\left(y\right)\right)=f\left(x,y\right)$
$\frac{\mathrm{\partial }}{\mathrm{\partial }y}\left(\frac{\mathrm{\partial }}{\mathrm{\partial }x}g\left(u,v\right)\right)=\frac{\mathrm{\partial }}{\mathrm{\partial }y}\left(\frac{\mathrm{\partial }g\left(u,v\right)}{\mathrm{\partial }u}\frac{\mathrm{\partial }u}{\mathrm{\partial }x}\right)=\frac{\mathrm{\partial }}{\mathrm{\partial }y}\left(\frac{\mathrm{\partial }g\left(u,v\right)}{\mathrm{\partial }u}\right)=\frac{{\mathrm{\partial }}^{2}g}{\mathrm{\partial }{u}^{2}}\frac{\mathrm{\partial }u}{\mathrm{\partial }y}+\frac{{\mathrm{\partial }}^{2}g}{\mathrm{\partial }u\phantom{\rule{thinmathspace}{0ex}}\mathrm{\partial }v}\frac{\mathrm{\partial }v}{\mathrm{\partial }y}=\phantom{\rule{0ex}{0ex}}\left(2y\frac{{\mathrm{\partial }}^{2}}{\mathrm{\partial }{u}^{2}}+2y\frac{{\mathrm{\partial }}^{2}}{\mathrm{\partial }u\phantom{\rule{thinmathspace}{0ex}}\mathrm{\partial }v}\right)g=2\sqrt{v}\left(\frac{{\mathrm{\partial }}^{2}}{\mathrm{\partial }{u}^{2}}+\frac{{\mathrm{\partial }}^{2}}{\mathrm{\partial }u\phantom{\rule{thinmathspace}{0ex}}\mathrm{\partial }v}\right)g.$
So we see that operator $\frac{{\mathrm{\partial }}^{2}}{\mathrm{\partial }x\mathrm{\partial }y}$ doesn't transform into $\frac{\mathrm{\partial }\left(u,v\right)}{\mathrm{\partial }\left(x,y\right)}\frac{{\mathrm{\partial }}^{2}}{\mathrm{\partial }u\mathrm{\partial }v}$ the same way as dxdy transforms to $\frac{\mathrm{\partial }\left(x,y\right)}{\mathrm{\partial }\left(u,v\right)}dudv$. Whereas in 1D $\frac{\mathrm{\partial }}{\mathrm{\partial }q}$ is indeed $\frac{\mathrm{\partial }p}{\mathrm{\partial }q}\frac{\mathrm{\partial }}{\mathrm{\partial }p}$. This is the reason why you can treat derivatives as fractions of differentials in 1D, but cannot do the same in 2D+.
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